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 Geometrical interpretation of Covector Fields 
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Joined: 28 Dec 2009, 21:25
Posts: 3
Post Re: Geometrical interpretation of Covector Fields
It's good to hear that I'm not the only one who is confused with this - probably I was just too used to all the grad(\phi)-things I'm doing right now for studies that I couldn't realize the subtle difference between gradient and grad-vector ;).

So Penrose uses gradient in the sense of a scalar, which is defined at a point and in direction of \xi, and we were used to the gradient as a vector at a point which points to max increase of \phi. Quite a confusion of definition, but I think I now got the idea of it. Seems that his way of looking at this is more general.

I'm really looking forward to the physics part of Penrose's book, waiting for all those technical terms to become useful.


31 Dec 2009, 15:11

Joined: 28 Oct 2009, 19:06
Posts: 7
Post Re: Geometrical interpretation of Covector Fields
Since my previous postings on this subject I've been re-reading chapters 10 and 12 again and again to try to get to what Penrose is saying.

I think I've finally understood the thing about the geometry of co-vectors. In diagram 10.8 (a) Penrose shows that dPhi points along the contour line ie the line of constant phi.

Previously I was trying to reconcile this with my "vector calculus" view of gradient whereby the gradient of a scalar field is in the line of steepest ascent, so that gradphi (or [unparseable or potentially dangerous latex formula] as referring to the plane (in a 3-manifold) spanned by all other coordinate axes other than the x^r axis, ie dx^r is a 2-dimensional (ie n-1 dimensional, for n=3) plane of constant x^r, specifically x^1 is constant over the plane spanned by the x^2 and x^3 axes.

Now go back to diagram 10.8. We have Phi constant along a contour line so this has to represent the covector dPhi. This is then consistent with the fact that Phi is actually increasing in a direction orthogonal to dPhi, compare with x^1 increasing in the direction orthogonal to the plane spanned by x^2 and x^3.

It's like the vector gradient grad Phi is orthogonal to the covector dPhi, so I suppose in this instance the vector gradient and the covector are duals.


07 Mar 2010, 19:16
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Joined: 07 Jun 2008, 08:21
Posts: 235
Post Re: Geometrical interpretation of Covector Fields
Just corrected the Latex:
dp4947 wrote:
Since my previous postings on this subject I've been re-reading chapters 10 and 12 again and again to try to get to what Penrose is saying.

I think I've finally understood the thing about the geometry of co-vectors. In diagram 10.8 (a) Penrose shows that d\Phi points along the contour line ie the line of constant \Phi.

Previously I was trying to reconcile this with my "vector calculus" view of gradient whereby the gradient of a scalar field is in the line of steepest ascent, so that grad\phi (or d\Phi) would in fact be orthogonal to the contours.

However, in diagram 12.9 (b) Penrose defines a covector dx^r as referring to the plane (in a 3-manifold) spanned by all other coordinate axes other than the x^r axis, ie dx^r is a 2-dimensional (ie n-1 dimensional, for n=3) plane of constant x^r, specifically x^1 is constant over the plane spanned by the x^2 and x^3 axes.

Now go back to diagram 10.8. We have \Phi constant along a contour line so this has to represent the covector d\Phi. This is then consistent with the fact that \Phi is actually increasing in a direction orthogonal to d\Phi, compare with x^1 increasing in the direction orthogonal to the plane spanned by x^2 and x^3.

It's like the vector gradient grad \Phi is orthogonal to the covector d\Phi, so I suppose in this instance the vector gradient and the covector are duals.


07 Mar 2010, 22:17

Joined: 28 Oct 2009, 19:06
Posts: 7
Post Re: Geometrical interpretation of Covector Fields
Thanks Vasco, not sure what I did wrong, but I'll get the hang of Latex one day !


08 Mar 2010, 16:41

Joined: 26 Mar 2010, 04:39
Posts: 109
Post Re: Geometrical interpretation of Covector Fields
Note that if \Phi is a scalar field, then
d\Phi = {\partial \Phi\over \partial x^i}dx^i = (\nabla \Phi)_i dx^i

You can see that d\Phi has the same components as what you would be used to calling the gradient vector from undergraduate vector calculus classes (and incidentally note the consistency in the notation of using the symbol \nabla to represent the covariant derivative). It is common to identify (\nabla \Phi)_i dx^i with (\nabla \Phi)_i e^i where e^i are the usual basis vectors in Euclidean space. Often in undergraduate classes they will even tell you that the gradient is a vector field, where in fact it is a covector field.

Remember a covector is just the dual of a vector where the dual space refers to the vector space of linear functionals on a vector space. In our case the vector space in question is the tangent space (strictly speaking we should say the tangent space evaluated at a particular point in a manifold) which has a basis
\left{{\partial\over\partial x^1},\dots {\partial\over\partial x^n}\right}
The dual space (covector space) is the cotangent space (again we should actually be evaluating at a given point) with basis
\left{ d x^1,\dots d x^n\right}
where we define the covector basis according to:
d x^i({\partial\over\partial x^j}) = \delta^i_j

If you want to try to think geometrically, then a covector field is something which allows you to assign a numerical value to each point in a vector field - in other words it acts on the vector field to produce a scalar field.

I hope this helps to answer your question.


20 Apr 2010, 17:17

Joined: 07 May 2010, 03:08
Posts: 1
Post Re: Geometrical interpretation of Covector Fields
Hi.. I am still confused even after all this discussion. In fig 10.8 penrose shows a scalar field defined on a 2-manifold S. In 12.3 page 225, he writes alpha determines a (n-1)-plane element. In the particular case when alpha is the gradient d(phi) the (n-1)-plane elements will be tangential to the contours.

What plane elements is he referring to? alpha (when its the gradient vector) itself is actually a 1-dimensional plane, orthogonal to the contour lines on our 2-manifold. If I am to think of plane elements as the basis elements of alpha, then they might be parallel at some point of the contour but won't be parallel at the point at which @ is being considered. As @ and kii (he symbol like E) are orthogonal, there wont even be a component of @ along kii. What am I to make of this?

Also, I ve difficulty visualizing @ when its not a gradient vector (d(phi)). I dont understand the 'twisting' diagram in fig. 12.8. :(

Any help wud be well appreciated!!


07 May 2010, 05:03

Joined: 26 Mar 2010, 04:39
Posts: 109
Post Re: Geometrical interpretation of Covector Fields
A hyperplane can be specified as either the span of n-1 vectors tangent to it or eqivalently by one normal.

Note that the space of p-forms has the same dimension as the space of (n-p)-forms and they can be treated as duals. In particular the space of 1-forms and the space of (n-1)-forms both have dimension n. The 1-forms (covectors) can be thought of as normal elements, while the corresponding (n-1)-forms are hyperplane volume elements.


07 May 2010, 07:11
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